Problem: If $x \oplus y = xy+3x-y$ and $x \otimes y = 6x+y$, find $3 \otimes (6 \oplus -1)$.
Solution: First, find $6 \oplus -1$ $ 6 \oplus -1 = (6)(-1)+(3)(6)-(-1)$ $ \hphantom{6 \oplus -1} = 13$ Now, find $3 \otimes 13$ $ 3 \otimes 13 = (6)(3)+13$ $ \hphantom{3 \otimes 13} = 31$.